\newproblem{lay:3_3_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.3.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Use the concept of volume to explain why the determinant of a $3\times 3$ matrix is zero iff $A$ is not invertible. 
}{
   % Solution
	From the invertible matrix theorem, we know that a matrix is invertible iff its columns are linearly independent. So the statement of this problem can be restated as
	the determinant of a $3\times 3$ matrix is zero iff the three columns of $A$ are linearly dependent. On the other side interpreting the determinant of $A$ as the volume
	of the parallelepiped formed by the three columns, the problem is ``the volume of the parallelepiped formed by three vectors is zero iff the three columns of $A$ are linearly
	dependent''.
	
	If the three vectors are linearly dependent, they span a subspace of dimension 2 or 1. In both cases, there is no real parallelepiped but a parallelogram or a segment and the volume
	of the parallelepiped is 0.
	
	Let us show that if the volume of the parallepiped is zero, then three columns are linearly dependent. Let's assume they are linearly independent. Then, they would actually
	span a three-dimensiional space, and the volume of the parallelepiped formed by the three would not be zero. But this is a contradiction with our hypothesis. So the three
	vectors have to be linearly dependent.
}
\useproblem{lay:3_3_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
